An equilateral triangular prism is made of glass `(mu = 1.5)`. A ray of light is incident normally on one of the faces. The angle between the incident and emergent ray is

Options

A. `60^(@)`
B. `90^(@)`
C. `120^(@)`
D. `180^(@)`

Answer

To find the angle between the incident and emergent ray when a ray of light passes through an equilateral triangular prism made of glass (with a refractive index \( \mu = 1.5 \)), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Prism and Incident Ray**:
- An equilateral triangular prism has angles of \( 60^\circ \) each.
- A ray of light is incident normally on one of the faces of the prism. This means the angle of incidence \( i_1 = 0^\circ \) (since it is normal incidence).

2. **Refraction at the First Face**:
- Since the ray is incident normally, it will pass through the first face without any deviation. Therefore, the angle of refraction \( r_1 = 0^\circ \).

3. **Determine the Angle of Incidence at the Second Face**:
- As the ray travels through the prism, it reaches the second face. The angle of incidence \( i_2 \) at the second face can be calculated using the geometry of the prism.
- The angle of the prism \( A = 60^\circ \).
- The angle of incidence at the second face \( i_2 \) can be found as:
\[ i_2 = 60^\circ - r_1 = 60^\circ - 0^\circ = 60^\circ \]

4. **Applying Snell's Law at the Second Face**:
- According to Snell's Law:
\[ \mu_1 \sin(i_2) = \mu_2 \sin(r_2) \]
- Here, \( \mu_1 = 1.5 \) (for glass), \( \mu_2 = 1.0 \) (for air), and \( i_2 = 60^\circ \).
- Plugging in the values:
\[ 1.5 \sin(60^\circ) = 1.0 \sin(r_2) \]
- Since \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \):
\[ 1.5 \cdot \frac{\sqrt{3}}{2} = \sin(r_2) \]
- Therefore:
\[ \sin(r_2) = \frac{1.5\sqrt{3}}{2} \]

5. **Finding the Angle of Refraction \( r_2 \)**:
- To find \( r_2 \), we need to calculate:
\[ r_2 = \arcsin\left(\frac{1.5\sqrt{3}}{2}\right) \]
- Since \( \frac{1.5\sqrt{3}}{2} \) is greater than 1, this indicates that total internal reflection occurs.

6. **Total Internal Reflection**:
- Since the angle of incidence \( i_2 = 60^\circ \) is greater than the critical angle for glass to air, the light will undergo total internal reflection.

7. **Emergent Ray**:
- After total internal reflection, the light will exit the prism at the first face again. The angle of emergence \( e \) will also be \( 60^\circ \) due to the symmetry of the prism.

8. **Calculating the Angle Between Incident and Emergent Ray**:
- The angle between the incident ray and the emergent ray can be calculated as:
\[ \text{Angle between incident and emergent ray} = i_2 + e = 60^\circ + 60^\circ = 120^\circ \]

### Final Answer:
The angle between the incident and emergent ray is \( 60^\circ \).